# Why has the Si diode been open circuited?

Why has the Si diode been open circuited? • what is your reasoning?> What do you know about the properties.? – Tony Stewart Sunnyskyguy EE75 Nov 12 '16 at 7:52
• What is the relationship of the voltage drops over two parallel components (here, the diodes)? What is the actual voltage drop? What does this imply for the current through the Si diode? – CL. Nov 12 '16 at 8:08
• @TonyStewart.EEsince'75 I know that voltage drop across a Si diode in forward biased condition is 0.7 V. So why has the Si diode(which is supposed to have a voltage drop of 0.7 V), have been open circuited and and why the Ge diode short circuited? If they had been connected in series the combined voltage drop offered would have been 0.7 + 0.3 = 1V. But what has to be done in case they are connected parallel as in the above case?and why? – Anwesa Roy Nov 12 '16 at 8:21
• @CL. Sir, thats exactly what I want to know. What to do in case there are different voltage drops across two devices when connected in parallel?and why? – Anwesa Roy Nov 12 '16 at 8:24
• @TonyStewart.EEsince'75 Sorry sir, Ge diode is not short, it offers a voltge drop. – Anwesa Roy Nov 12 '16 at 8:28

Here is a graph showing a germanium and silicon diode current versus voltage graph; Let's consider if you only had the germanium diode in series with the 2.2k Ohm resistor for now. One way to solve for the voltage is to do it graphically. This is done by plotting the load line of the resistor. This load line is ploted by shorting the diode to and finding the current through the resistor is 5.5mA. To find another point on the graph the diode is open circuited and the voltage is found to be 12V. The line through (0V,5.5ma) and (12V,0mA) can be plotted plotted on this graph. The intersection of this line and the diode curve is this diodes opperating point.

Now you should see the Ge diode limits the voltage to a little more than 0.3V. So now if you put the Si diode back in the circuit, it can't have more than this 0.3V across it. The Si diode at 0.3V has zero current through it.

The answer to you question is that the Si diode has been open circuited because the Ge diode limits the voltage across it to 0.3 where it has zero current.

In finding the answer for your question look at the properties of the diodes. Maybe you know the difference between a germanium and silicium diode.

If not then try to find information concerning the voltage- current caracteristic of both diodes.

You can find this information in Wikipedia or the datasheets of germanium and silicium diodes.

Once you have this you should be able to answer the question yourself.